bnu 34986 Football on Table(數學+暴力)

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題目串連：bnu 34986 Football on Table

題目大意:給出桌子的大小L，W，然後是球的起始位置sx，sy，以及移動的向量dx，dy，然後給出n，表示有n個杆，對於每個杆，先給出位置x，以及杆上有多少個小人c，給出小人的寬度，再給出c個小人間的距離。現在問說球有多少個機率可以串過所有人。

解題思路；對於每個杆求無阻擋的機率，注意機率 = 空隙 / 可移動的範圍大小，而不是W。其他就水水的。

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const double eps = 1e-8;const int N = 205;double L, W;double sx, sy, dx, dy;double d[N];double solve () {    int n, c;    double w, ans = 1;    scanf("%d", &n);    while (n--) {        scanf("%lf%d", &w, &c);        double r = sy + w * dy / dx;        double s = 0;        c = 2 * c - 1;        for (int i = 0; i < c; i += 2)            scanf("%lf", &d[i]);        for (int i = 1; i < c; i += 2)            scanf("%lf", &d[i]);            for (int i = 0; i < c; i++)            s += d[i];        double l = W - r;        double rec = 0;        if (l > s) {            rec += (l - s);            l = 0;        } else {            l = s - l;        }        if (r > s) {            rec += (r - s);            r = s;        }        s = 0;        for (int i = 0; i < c; i++) {            double tmp = s + d[i];            if (i&1) {                double add = min(r, tmp) - max(s, l);                rec += max(add, (double)0);            }            s = tmp;        }        ans *= rec / (W-s);    }    return ans;}int main () {    int cas;    scanf("%d", &cas);    for (int i = 1; i <= cas; i++) {        scanf("%lf%lf", &L, &W);        scanf("%lf%lf%lf%lf", &sx, &sy, &dx, &dy);        printf("Case #%d: %.5lf\n", i, solve());    }    return 0;}